Timeline for A priori estimate of an inhomogeneous p-Laplace equation with Dirichlet boundary condition
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 17, 2017 at 23:40 | comment | added | fedja | @ANZM91 Yes, there is a constant missing :-) | |
| Dec 13, 2017 at 18:24 | comment | added | ANZM91 | hello fedja, my setup is $\Omega \subset R^2$ so everything good. but another question came to my mind: why we can easy say that \begin{align} ||f||_{L^2} ||u - g||_{L^2} \leq ||f||_{L^2} ||u - g||_{W^{1,p}} \end{align} isn't there a constant missing? or is $||u - g||_{L^2} \leq ||u - g||_{W^{1,p}}$? At least as far as i know $||u - g||_{L^2} \geq ||u - g||_{L^p}$ | |
| Dec 12, 2017 at 19:12 | comment | added | fedja | @ANZM91 That would require $f\in L^{p/(p-1)}$. If it is there, you are fine. However, $W_{1,p}\subset L^2$ if $\frac 1p\le \frac 12+\frac 1n$ ($n$ is the dimension of the space and we assume that $\Omega$ is bounded), so you have some leeway here (all the way up to $p=1$ in $\mathbb R^2$). So, you have to check your setup to see if $p$ is within the required range or not to use just the $L^2$ bound for $f$. | |
| Dec 12, 2017 at 17:35 | comment | added | ANZM91 | Ah ok! And thank you again!!! :) You can't believe how grateful I am for your fast and good help! But regarding my with "Assuming $W^{1,p} \subset L^2$: In my case I have $p \geq 1$ and for $p<2$ it doesnt work, does it? What do I have to change? Do I have to estimate this way using hoelder inequality? \begin{align*} \int_{\Omega} f (u-g) \leq ||f||_{p/(p-1)} ||u - g||_p \end{align*} | |
| Dec 12, 2017 at 17:25 | comment | added | fedja | @ANZM91 Ah, yes. You used the condition that the difference is in $W^{1,p}_0$ and I was thinking of $W^{1,p}$. Then you are right here and everything works as you said. As to the other question, $4=9$ for all practical purposes (i.e. they are the same up to a constant factor). | |
| Dec 12, 2017 at 17:09 | comment | added | ANZM91 | Thank you again for your help! I hope I dont annoy you, but how can $(a + b)^2 = a^2 + b^2$ be true? $(1+2)^2 = 9 \not= 1^2 + 2^2 = 4$ ? I feel a bit dumb know, can't get the point >.< And why is the first inequality wrong? I just used the Poincaé inequality because $u - g \in W^{1,p}_0$... And I again don't know what you mean with $ u - g = 1$ (constantly 1 nearly everywhere on $\Omega$? | |
| Dec 12, 2017 at 16:57 | comment | added | fedja | @ANZM91 The very first inequality in your chain is wrong: think of $u-g\equiv 1$ | |
| Dec 12, 2017 at 16:53 | comment | added | ANZM91 | Ahhhh, maybe you did mean this way? \begin{align*} ||u - g||^p_{1,p} &\leq C_p || \nabla u - \nabla g ||^p_{p} \\ &\leq C_p(||\nabla u||_p + ||\nabla g||_p)^p \\ &= C_p \sum_{i=0}^{p} \binom{p}{i} ||\nabla u||_p^{p - i}||\nabla g||_p^i \\ &\leq C_p \binom{p}{p/2} (||\nabla u||^p_p + ||\nabla g||^p_p) \\ &\leq C (||\nabla u||^p_p + ||\nabla g||^p_p + ||\nabla g||) \\ &= \leq C (||\nabla u||^p_p + ||\nabla g||^p_{1,p}) \end{align*} By the way: In my case I have $p\geq 1$ so $W^{1,p} \subset L^2$ isnt working, is it? What do i have to change then? | |
| Dec 12, 2017 at 16:50 | comment | added | fedja | @ANZM91 Analysis is not algebra. If $a,b,p>0$, then we have $(a+b)^p=a^p+b^p$ for all practical purposes. My RHS is larger because it also includes $\|g\|_p^p$, which you lost somewhere (and it matters!). | |
| Dec 12, 2017 at 14:49 | comment | added | ANZM91 | Why is it larger? I have $(||\nabla u||_p + ||g||_p + ||\nabla g||_p ) ^p$ - the p-th power of the sum. Or is $| x - y| ^p \leq |x|^p + |y|^p $ always true? Sorry, but I am a bit confused at the moment >.< | |
| Dec 12, 2017 at 14:25 | comment | added | fedja | @ANZM91 What do you mean? $\|g\|_{1.p}^p\approx \|g\|_p^p+\|\nabla g\|_p^p$, so my RHS is larger than yours (and $\|g\|_p$ matters, by the way: think of large constants). | |
| Dec 12, 2017 at 13:29 | comment | added | ANZM91 | Wow! Thank you very very much! Yes, I am new too such estimates and this functional analysis stuff, so I am very gratefull for your detailed answer. But I still have one question regarding your mentioned bound $||u - g||^p_{1,p} \leq C [|| \nabla u||^p + ||g||^p_{1,p}]$... Can you explain why it is true? Because in my eyes it's not - at least not for arbitrary u and g? :/ I only have $||u - g||^p_{1,p} \leq C ||\nabla u - \nabla g||^p_p \leq C ( ||\nabla u||_p + || \nabla g ||_p)^p$ | |
| Dec 11, 2017 at 21:50 | history | answered | fedja | CC BY-SA 3.0 |